Dynamic Programming

Md. Shakil AhmedSoftware Engineer Astha it research & consultancy ltd.Dhaka, Bangladesh

Dynamic Programming• Algorithmic technique that systematically records the answers to sub-problems in a table and re-uses those recorded results (rather than re-computing them).• Simple Example: Calculating the Nth Fibonacci number.

Fib(N) = Fib(N-1) + Fib(N-2)

Introduction

Topic Focus:

• Unidirectional TSP• Coin Change • LCS• LIS• The Partition Problem

• String Partition• TSP (Bitmask)

• Matrix-chain multiplication

Unidirectional TSP

Input:5 63 4 1 2 8 66 1 8 2 7 45 9 3 9 9 58 4 1 3 2 63 7 2 8 6 4

Output:1 2 3 4 4 516

Output minimal-weight path, and the second line is the cost of a minimal path.

Source Codefix= infinity;for(i=0;i<n;i++) for(j=0;j<m;j++) { scanf("%lld",&sa[i][j]); si[i][j]=fix; t1[i][j]=-1; } for(i=0;i<n;i++) si[i][m-1]=sa[i][m-1];

for(j=m-1;j>0;j--) for(i=0;i<n;i++) { if(si[i][j-1] > si[i][j] + sa[i][j-1] || (si[i][j-1] == si[i][j] + sa[i][j-1] && t1[i][j-1] > i)) { si[i][j-1]=si[i][j]+sa[i][j-1]; t1[i][j-1]=i; } k=i-1; if(k<0) k=k+n; if(si[k][j-1] > si[i][j] + sa[k][j-1] || (si[k][j-1] == si[i][j] + sa[k][j-1] && t1[k][j-1] > i)) {

si[k][j-1]=si[i][j]+sa[k][j-1];t1[k][j-1]=i;

}

Source Codek=i+1;k=k%n; if(si[k][j-1] > si[i][j] + sa[k][j-1] || (si[k]

[j-1] == si[i][j] + sa[k][j-1] && t1[k][j-1] > i))

{si[k][j-1]=si[i][j]+sa[k][j-1];t1[k][j-1]=i;

} }min=fix; for(i=0;i<n;i++) if(si[i][0]<min) {min=si[i][0];k=i;}

printf("%ld",k+1);

j=0;

while(t1[k][j]!=-1) { printf(" %ld",t1[k][j]+1); k=t1[k][j]; j++; }

printf("\n%lld\n",min);

Sample Problems:•UVA Online Judge: 116.

Coin change

• Coin change is the problem of finding the number of ways to make change for a target amount given a set of denominations. It is assumed that there is an unlimited supply of coins for each denomination. An example will be finding change for target amount 4 using change of 1,2,3 for which the solutions are (1,1,1,1), (2,2), (1,1,2), (1,3).

Coin Change Source Codelong s[6], sa[7600]={0},

i, j, k;

s[0]=1;

s[1]=5;

s[2]=10;

s[3]=25;

s[4]=50;

sa[0]=1;

for(j=0;j<=4;j++){ for(i=0;i<=7500;i++) { if(i+s[j]>7500) break;

sa[i+s[j]]+=sa[i]; } } while(scanf("%ld",&i)==1){ printf("%ld\n",sa[i]);}

Coin change

• Sample Problems: 1.UVA Online Judge: 147, 166, 357, 674, 10306,

10313, 11137, 11517.

The longest common subsequence (LCS) problem

• A string : A = b a c a d• A subsequence of A: deleting 0 or more symbols

from A (not necessarily consecutive).e.g. ad, ac, bac, acad, bacad, bcd.• Common subsequences of A = b a c a d and B = a c c b a d c b : ad, ac, bac, acad.• The longest common subsequence (LCS) of A and

B: a c a d.

The LCS algorithm

• Let A = a1 a2 am and B = b1 b2 bn

• Let Li,j denote the length of the longest common subsequence of a1 a2 ai and b1 b2 bj.

•  Li,j = Li-1,j-1 + 1 if ai=bj

max{ Li-1,j, Li,j-1 } if aibj

L0,0 = L0,j = Li,0 = 0 for 1im, 1jn.

• Time complexity: O(mn)• The dynamic programming approach for solving

the LCS problem:L1,1

L2,1

L3,1

L1,2 L1,3

L2,2

Lm,n

Tracing back in the LCS algorithm• e.g. A = b a c a d, B = a c c b a d c b

• After all Li,j’s have been found, we can trace back to find the longest common subsequence of A and B.

2

43

100000

0 0 0

111

1 1 11 1 1 1 1

22 2 2

222

2 2 2 222

33 3 3

33

4 4

0 0 0 0 0 0 0 00 0 0 0 0

bac

a

d

A

ac c ba d cbB

2

LCS Source Codevoid LCS(){ for(i=0;i<=m;i++) for(j=0;j<=n;j++) {

c[i][j]=0;b[i][j]='0';

} for(i=1;i<=m;i++) for(j=1;j<=n;j++) {

if(x[i]==y[j]) {

c[i][j]=c[i-1][j-1]+1; b[i][j]='1';

}

else if(c[i-1][j]>=c[i][j-1]) {

c[i][j]=c[i-1][j]; b[i][j]='2';

} else { c[i][j]=c[i][j-1]; b[i][j]='3'; } }}

void printSequence(long int i,long int j){

if(i==0||j==0) return ; if(b[i][j]=='1') {

printSequence (i-1,j-1); printf(“%c”,x[i]); } else if(b[i][j]=='2') printSequence (i-1,j); else printSequence (i,j-1);}

Sample Problems:• UVA Online Judge: 10066,

10405.

Longest Increasing Subsequence• The longest increasing subsequence problem is to find a

subsequence of a given sequence in which the subsequence elements are in sorted order, lowest to highest, and in which the subsequence is as long as possible. This subsequence is not necessarily contiguous.

• A simple way of finding the longest increasing subsequence is to use the Longest Common Subsequence (Dynamic Programming) algorithm.

1. Make a sorted copy of the sequence A, denoted as B. O(nlog(n)) time.

2. Use Longest Common Subsequence on with A and B. O(n2) time.

Simple LIS

function lis_length( a ) n := a.length q := new Array(n) for k from 0 to n:

max := 0; for j from 0 to k, if a[k] > a[j]:

if q[j] > max, then set max = q[j]. q[k] := max + 1;

max := 0 for i from 0 to n:

if q[i] > max, then set max = q[i]. return max;

Simple LIS• Complexity O(n^2)for(i=0;i<n;i++){

scanf(“%ld”,&A[i]);Len[i]=1;Pre = -1;

}Max = 0; Position = 0;for(i=0;i<n;i++){

if(Len[i]>Max){

Max = Len[i];Position = i;

}

for(j=i+1;j<n;j++) If(A[j]>=A[i] && Len[j]<Len[i]+1)

{Len[j]=Len[i]+1;Pre[j]=i;}

}For sequence Print:void Show(long h1){

If(Pre[h1]!=-1)Show(Pre[h1]);

Printf(“%ld\n”,A[h1]);}

Show(Position);

Efficient LIS• Complexity O(n*log(n))L = 0 for i = 1, 2, ... n:

binary search for the largest positive j ≤ L such that X[M[j]] < X[i] (or set j = 0 if no such value exists)

P[i] = M[j] if j == L or X[i] < X[M[j+1]]:

M[j+1] = i L = max(L, j+1)

Efficient LISlong BinarySearch(long left,long right,long value){long mid = (left+right)/2;while(left<=right){

if(temp[mid]==value)break;

else if(temp[mid]<value)left = mid+1;

elseright = mid – 1;

mid = (left+right)/2;}return mid;}

Efficient LISvoid LIS(long N){ long n=0,i,mid;

for(i=0;i<N;i++){if(n==0){ temp[n]=sa[i];tempP[n]=i;

A[i]=1;n++; }else{mid = BinarySearch(0,n-1,sa[i]);while(mid>0&&temp[mid]>sa[i])

mid–;if(temp[mid]<sa[i]){

A[i]=mid+2;P[i]=tempP[mid];

if(mid+1>=n){temp[n]=sa[i]; tempP[n]=i;n++;}else if(temp[mid+1]>sa[i]){temp[mid+1]=sa[i];tempP[mid+1]=i;}

}else if(temp[0]>sa[i]){temp[0]=sa[i];tempP[0]=i;} } }}

Efficient LISvoid show(long h1)

{if(P[h1]==-1)printf(“%ld”,sa[h1]);else{show(P[h1]);printf(” %ld”,sa[h1]);}}

for(i=0;i<n;i++) {scanf(“%ld”,&sa[i]);P[i]=-1;}LIS(n);max = 0;x = 0;for(i=0;i<n;i++)if(A[i]>max){max = A[i];x = i;}printf(“%ld\n”,max);show(x);

LIS

• Sample Problems:1.UVA Online Judge: 111, 231, 437, 481, 497,

1196, 10131, 10534, 11456, 11790.

Given a set of positive integers, A = {a1, a2, …, an}. The question is to select a subset B of A such that the sum of the numbers in B equals the sum of the numbers not in B, i.e., . We may assume that the sum of all numbers in A is 2K, an even number. We now propose a dynamic programming solution. For 1 i n and 0 j K,

define P[i, j] = True if there exists a subset of the first i numbers a1 through ai whose sum equals j; False otherwise.

Thus, P[i, j] = True if either j = 0 or if (i = 1 and j = a1). When i > 1, we have the following recurrence: P[i, j] = P[i – 1, j] or (P[i – 1, j – ai] if j – ai 0) That is, in order for P[i, j] to be true, either there exists a subset of the first i – 1 numbers whose sum equals j, or whose sum equals j – ai (this latter case would use the solution of P[i – 1, j – ai] and add the number ai. The value P[n, K] is the answer.

BAa

jBai

ji

aa

The Partition Problem

Example: Suppose A = {2, 1, 1, 3, 5} contains 5 positive integers. The sum of these number is 2+1+1+3+5=12, an even number. The partition problem computes the truth value of P[5, 6] using a tabular approach as follows:

a1 = 2 a2 = 1 a3 = 1 a4 = 3 a5 = 5

0 1 2 3 4 5 6

T F T F F F F T T T T F F F T T T T T F F T T T T T T T T T T T T T T

ji

Always true

P[4,5] = (P[3,5] or P[3, 5 – a4] )

The time complexity is O(nK); the space complexity O(K).

Because j a1

The Partition Problem Source Code

sum = 0;for(i=0;i<n;i++)

sum += A[i];

for(i=0;i<=sum/2;i++)Flag[i]=0;

Flag[0]=1;

for(i=0;i<n;i++)for(j=sum/2-A[i];j>=0;j--)if(Flag[j]==1)

Flag[j+A[i]]=1;

for(i=sum/2;i>=0;i--)if(Flag[i]==1)

break;

Dif = (sum – i) – i;

String Partition

A string of digits instead of a list of integers. There are many ways to split a string of digits into a list of non-zero-leading (0 itself is allowed) 32-bit signed integers. What is the maximum sum of the resultant integers if the string is split appropriately? A string of at most 200 digits.

Input:11111111111111111111111111111111111111111111111111111Output:5555555666

Source CodeLimit = 2147483647;Len = strlen(input);for(i=0;i<=Len;i++)

Max[i]=0;for(i=0;i<Len;i++){

Num = 0;for(j=i;j<Len;j++){

Num = Num * 10 + input[j]-’0’;if(Num>Limit)

break;if(Num+Max[i]>Max[j+1]) Max[j+1] =

Num+Max[i];

if(j==i&&input[j]==‘0’)break;

}}

printf(“%lld\n”,Max[Len]);

Sample Problems:•UVA Online Judge: 11258.

The traveling salesperson (TSP) problem

• e.g. a directed graph :

• Cost matrix:

12

3

2

44

2

56

7

104

8

39

1 2 3 4

1 2 10 5 2 2 9 3 4 3 4 4 6 8 7

• A multistage graph can describe all possible tours of a directed graph.

• Find the shortest path:(1, 4, 3, 2, 1) 5+7+3+2=17

(1) (1,3)

(1,2)

(1,4)

2

5

10

(1,2,3)

(1,2,4)

(1,3,2)

(1,3,4)

(1,4,2)

(1,4,3)

9

3

4

8

7

¡Û

(1,2,3,4)

(1,2,4,3)

(1,3,2,4)

(1,3,4,2)

(1,4,2,3)

(1,4,3,2)

¡Û

4

7

8

9

3

1

4

6

6

2

4

2

TSP

BIT MASK• you are in node 1, and you have to visit node

2 5 & 10. Now, you have to fine the minimum cost for this operation.

• you have to visit l number of node, like l = 3. then node[] = 2, 3, 4& your source is 1. now you have to find the minimum cost of the tour.

• C[][] = cost matrix of the nodes.

BIT MASK Source CodeP[0]=1;for(i=1;i<=15;i++ ){

P[i] = P[i-1]*2;}

source = 1;

for(i=0;i<=P[l];i++){

for(j=0;j<l;j++){

bit[i][j] = MAX;}

}

for(i=0;i<l;i++){ if( C[ source ][ node[i] ]!=MAX ) { bit[ P[i] ][i] = C[ source ][ node[i] ]; }}for(i=0;i<P[l];i++){ for(j=0;j<l;j++) { if( bit[i][j]!=MAX ) { for(k=0;k<l;k++) { if( (i%P[k+1])/P[k]==0 ) {

BIT MASK Source Code

if( C[node[j]][node[k]]!=MAX ){

if( bit[ i+P[k] ][k] > bit[i][j] + C[node[j]][node[k]] ){

bit[ i+P[k] ][k] = bit[i][j] + C[node[j]][node[k]];}

}}}}}}

int res = MAX;

for(i=0;i<l;i++) if(bit[P[l]-1][i]!=MAX) if(res > bit[P[l]-1][i]) res = bit[P[l]-1][i];

printf(“%d\n”,res);

Sample

• UVA Online Judge: 10651, 216, 10496, 11813, 10718.

Matrix-chain multiplication• n matrices A1, A2, …, An with size p0 p1, p1 p2, p2 p3, …, pn-1 pn

To determine the multiplication order such that # of scalar multiplications is minimized.

• To compute Ai Ai+1, we need pi-1pipi+1 scalar multiplications.

e.g. n=4, A1: 3 5, A2: 5 4, A3: 4 2, A4: 2 5((A1 A2) A3) A4, # of scalar multiplications:

3 * 5 * 4 + 3 * 4 * 2 + 3 * 2 * 5 = 114(A1 (A2 A3)) A4, # of scalar multiplications:

3 * 5 * 2 + 5 * 4 * 2 + 3 * 2 * 5 = 100(A1 A2) (A3 A4), # of scalar multiplications:

3 * 5 * 4 + 3 * 4 * 5 + 4 * 2 * 5 = 160

• Let m(i, j) denote the minimum cost for computing Ai Ai+1 … Aj

• Time complexity : O(n3)• Computation sequence :

jiif

jiifpppj)1,m(kk)m(i,min

0j)m(i,

ik1ijki

Matrix-chain multiplication// Matrix Ai has dimension p[i-1] x p[i] for i = 1..n Matrix-Chain-Order(int p[]) { // length[p] = n + 1 n = p.length - 1; // m[i,j] = Minimum number of scalar multiplications (i.e., cost) // needed to compute the matrix A[i]A[i+1]...A[j] = A[i..j] // cost is zero when multiplying one matrix for (i = 1; i <= n; i++)

m[i,i] = 0;

for (L=2; L<=n; L++) { // L is chain length for (i=1; i<=n-L+1; i++) {

j = i+L-1;

Matrix-chain multiplicationm[i,j] = MAXINT; for (k=i; k<=j-1; k++) {

// q = cost/scalar multiplications q = m[i,k] + m[k+1,j] + p[i-1]*p[k]*p[j]; if (q < m[i,j]) {

m[i,j] = q; // s[i,j] = Second auxiliary table that stores k // k = Index that achieved optimal cost s[i,j] = k;

} } } } }

Thanks!